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We denote by $A$, $A′$, $B$, $B′$ the four sets.

We are given mapping $u:A′→A$, $v:B→B′$.

We denote by $F(A, B)$ the set maps $A→B$ ，$F(A′, B′)$ the set maps $A′→B′$.

Let $Φ : F(A, B) → F(A′, B′)$ be a mappig which makes $v◦f ◦ u ∈ F(A′, B′)$ correspond with for $f ∈ F(A, B)$.

Prove that we can't prove "$u$ is injective and $v$ is surjective $⇒$ $Φ$ is surjective." without Axiom of choice.

md2perpe
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Pacifica
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  • I'm confused. Are we to show the validity of your implication implies the axiom of choice? – AlvinL Jul 15 '18 at 08:49
  • I just want to show that

    it's impossible to prove "$u$ is injective and $v$ is surjective $⇒$ $Φ$ is surjective" without using Axiom of choice.

     – Pacifica Jul 15 '18 at 08:57
  • So you want to show the undecidability of this implication in ZF, right? – joriki Jul 15 '18 at 09:28
  • @AlvinLepik : essentially OP wants a choiceless model with $2$ maps $u,v$ that are injective (resp. surjective) such that $f\mapsto v\circ f\circ u$ is not surjective – Maxime Ramzi Jul 15 '18 at 09:28
  • Either that or a proof of AC from the implication – Maxime Ramzi Jul 15 '18 at 09:29
  • I mean, I could't come up with a proof without the use of axiom of choice, and I think we can't do that, I want to show that. – Pacifica Jul 15 '18 at 09:35

1 Answers1

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Actually, the $u$ part is misleading, because $f\mapsto f\circ u$ is surjective if $u$ is injective, without the axiom of choice. The interesting part is about $v$ !

So we may as well take $u= id_{A'}$. And now the implication we have is "$v$ surjective implies $f\mapsto v\circ f$ surjective" and we want to see that this implies the axiom of choice.

Consider : $f\mapsto v\circ f$, $F(B', B)\to F(B',B')$.

If you take $id_{B'}\in F(B',B')$, the surjectivity of this map implies that $v$ has a section, i.e. there is a function $B'\to B$ that picks out an antecedent for each element of $B'$.

It's known that "every surjective function has a section" implies the axiom of choice, so there you go

Maxime Ramzi
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  • Actually, there was another proposition: "$u$ is surjective and $v$ is injective $⇒$ $Φ$ is injective"...(ii)

    We can prove (ii) without the axiom of choice simply. So, without the axiom of choice: We can prove the question I posted, but can't prove (ii). So, I think this would be one of the reasons which vindicate the axiom of choice.

     – Pacifica Jul 15 '18 at 13:57